Lab Report Activation energy

[MWL]

In order for magnesium to react with oxygen, is there always a need for activation energy?

[Emmi]

For any reaction there will be an activation energy. Molecules just don't automatically react with one another to form a product when you put them together. In order for a successful reaction to occur, the molecules must have enough energy (and proper alignment) to overcome the energy barrier (the activation energy). Reactions with a low activation energy tend to, although not always, react faster than a reaction with a higher activation energy because the energy difference is not as high, and therefore more molecules will be able to react, which in turn increases the rate of your reaction. Typically you can overcome this by using an appropriate catalyst or increasing the temperature.

[MWL]

For any reaction there will be an activation energy. Molecules just don't automatically react with one another to form a product when you put them together. In order for a successful reaction to occur, the molecules must have enough energy (and proper alignment) to overcome the energy barrier (the activation energy). Reactions with a low activation energy tend to, although not always, react faster than a reaction with a higher activation energy because the energy difference is not as high, and therefore more molecules will be able to react, which in turn increases the rate of your reaction. Typically you can overcome this by using an appropriate catalyst or increasing the temperature.

And is it possible for the same element to have different activation energies once its solid shape changes from magnesium ribbon to pieces to powder? Because how am I supposed to prove that the surface area has an influence in the calculation of the magnesium oxide empirical formula? (based on experimental data: Mg₉O₅ (crucible A), Mg₁₉O₁₀ (crucible B) and Mg₁₈O (crucible D))

[Emmi]

For any reaction there will be an activation energy. Molecules just don't automatically react with one another to form a product when you put them together. In order for a successful reaction to occur, the molecules must have enough energy (and proper alignment) to overcome the energy barrier (the activation energy). Reactions with a low activation energy tend to, although not always, react faster than a reaction with a higher activation energy because the energy difference is not as high, and therefore more molecules will be able to react, which in turn increases the rate of your reaction. Typically you can overcome this by using an appropriate catalyst or increasing the temperature.

And is it possible for the same element to have different activation energies once its solid shape changes from magnesium ribbon to pieces to powder? Because how am I supposed to prove that the surface area has an influence in the calculation of the magnesium oxide empirical formula? (based on experimental data: Mg₉O₅ (crucible A), Mg₁₉O₁₀ (crucible B) and Mg₁₈O (crucible D))

You'll find elements in many different compounds (look at chlorine for example, you have HCl, NaCl, PCl5, etc) but the reason why there's no constant activation energy for chlorine is because the nature of each molecule is different (different sizes, geometries/shapes, bonds, etc). However, if your molecule is the same, changing the surface area will not lower the activation energy, it will just increase the rate of reaction. The reasoning behind this is because when you have a solid chunk of material, there isn't a lot of surface area to your material. However, if you took that solid chunk and crushed it into a fine powder, now you've got a lot of tiny little chunks that even though individually don't have a lot of surface area, adding up all of those tiny little chunks creates far more surface area. The surface area of a lot of little chunks is far greater than the surface area of one big chunk. Because there's more surface area, more molecules are able to come into contact with the other reactants, and because of that your reaction rate will increase. In short, more surface area = greater reaction surface = faster reaction.

With your specific reactants, what you're seeing is not a lowering of activation energy, you're seeing that changing the appearance is providing a greater surface area for the reaction which causes it to increase for the reasons stated above.

[MWL]

For any reaction there will be an activation energy. Molecules just don't automatically react with one another to form a product when you put them together. In order for a successful reaction to occur, the molecules must have enough energy (and proper alignment) to overcome the energy barrier (the activation energy). Reactions with a low activation energy tend to, although not always, react faster than a reaction with a higher activation energy because the energy difference is not as high, and therefore more molecules will be able to react, which in turn increases the rate of your reaction. Typically you can overcome this by using an appropriate catalyst or increasing the temperature.

And is it possible for the same element to have different activation energies once its solid shape changes from magnesium ribbon to pieces to powder? Because how am I supposed to prove that the surface area has an influence in the calculation of the magnesium oxide empirical formula? (based on experimental data: Mg₉O₅ (crucible A), Mg₁₉O₁₀ (crucible B) and Mg₁₈O (crucible D))

You'll find elements in many different compounds (look at chlorine for example, you have HCl, NaCl, PCl5, etc) but the reason why there's no constant activation energy for chlorine is because the nature of each molecule is different (different sizes, geometries/shapes, bonds, etc). However, if your molecule is the same, changing the surface area will not lower the activation energy, it will just increase the rate of reaction. The reasoning behind this is because when you have a solid chunk of material, there isn't a lot of surface area to your material. However, if you took that solid chunk and crushed it into a fine powder, now you've got a lot of tiny little chunks that even though individually don't have a lot of surface area, adding up all of those tiny little chunks creates far more surface area. The surface area of a lot of little chunks is far greater than the surface area of one big chunk. Because there's more surface area, more molecules are able to come into contact with the other reactants, and because of that your reaction rate will increase. In short, more surface area = greater reaction surface = faster reaction.

With your specific reactants, what you're seeing is not a lowering of activation energy, you're seeing that changing the appearance is providing a greater surface area for the reaction which causes it to increase for the reasons stated above.

however a faster reaction doesn't necessarily mean the result of the calculation for the magnesium oxide empirical formula will be more accurate, does it? in other words, the bigger the surface area the better the accuracy of the calculus?

[Emmi]

For any reaction there will be an activation energy. Molecules just don't automatically react with one another to form a product when you put them together. In order for a successful reaction to occur, the molecules must have enough energy (and proper alignment) to overcome the energy barrier (the activation energy). Reactions with a low activation energy tend to, although not always, react faster than a reaction with a higher activation energy because the energy difference is not as high, and therefore more molecules will be able to react, which in turn increases the rate of your reaction. Typically you can overcome this by using an appropriate catalyst or increasing the temperature.

And is it possible for the same element to have different activation energies once its solid shape changes from magnesium ribbon to pieces to powder? Because how am I supposed to prove that the surface area has an influence in the calculation of the magnesium oxide empirical formula? (based on experimental data: Mg₉O₅ (crucible A), Mg₁₉O₁₀ (crucible B) and Mg₁₈O (crucible D))

You'll find elements in many different compounds (look at chlorine for example, you have HCl, NaCl, PCl5, etc) but the reason why there's no constant activation energy for chlorine is because the nature of each molecule is different (different sizes, geometries/shapes, bonds, etc). However, if your molecule is the same, changing the surface area will not lower the activation energy, it will just increase the rate of reaction. The reasoning behind this is because when you have a solid chunk of material, there isn't a lot of surface area to your material. However, if you took that solid chunk and crushed it into a fine powder, now you've got a lot of tiny little chunks that even though individually don't have a lot of surface area, adding up all of those tiny little chunks creates far more surface area. The surface area of a lot of little chunks is far greater than the surface area of one big chunk. Because there's more surface area, more molecules are able to come into contact with the other reactants, and because of that your reaction rate will increase. In short, more surface area = greater reaction surface = faster reaction.

With your specific reactants, what you're seeing is not a lowering of activation energy, you're seeing that changing the appearance is providing a greater surface area for the reaction which causes it to increase for the reasons stated above.

however a faster reaction doesn't necessarily mean the result of the calculation for the magnesium oxide empirical formula will be more accurate, does it? in other words, the bigger the surface area the better the accuracy of the calculus?

It shouldn't affect the calculation. The rate shouldn't affect what the empirical formula is, because you can have the same molecule undergoing the same reaction but at different rates depending on external factors like temperature and surface area which don't change it. What calculations are you using to find the empirical formula?

[MWL]

For any reaction there will be an activation energy. Molecules just don't automatically react with one another to form a product when you put them together. In order for a successful reaction to occur, the molecules must have enough energy (and proper alignment) to overcome the energy barrier (the activation energy). Reactions with a low activation energy tend to, although not always, react faster than a reaction with a higher activation energy because the energy difference is not as high, and therefore more molecules will be able to react, which in turn increases the rate of your reaction. Typically you can overcome this by using an appropriate catalyst or increasing the temperature.

And is it possible for the same element to have different activation energies once its solid shape changes from magnesium ribbon to pieces to powder? Because how am I supposed to prove that the surface area has an influence in the calculation of the magnesium oxide empirical formula? (based on experimental data: Mg₉O₅ (crucible A), Mg₁₉O₁₀ (crucible B) and Mg₁₈O (crucible D))

You'll find elements in many different compounds (look at chlorine for example, you have HCl, NaCl, PCl5, etc) but the reason why there's no constant activation energy for chlorine is because the nature of each molecule is different (different sizes, geometries/shapes, bonds, etc). However, if your molecule is the same, changing the surface area will not lower the activation energy, it will just increase the rate of reaction. The reasoning behind this is because when you have a solid chunk of material, there isn't a lot of surface area to your material. However, if you took that solid chunk and crushed it into a fine powder, now you've got a lot of tiny little chunks that even though individually don't have a lot of surface area, adding up all of those tiny little chunks creates far more surface area. The surface area of a lot of little chunks is far greater than the surface area of one big chunk. Because there's more surface area, more molecules are able to come into contact with the other reactants, and because of that your reaction rate will increase. In short, more surface area = greater reaction surface = faster reaction.

With your specific reactants, what you're seeing is not a lowering of activation energy, you're seeing that changing the appearance is providing a greater surface area for the reaction which causes it to increase for the reasons stated above.

however a faster reaction doesn't necessarily mean the result of the calculation for the magnesium oxide empirical formula will be more accurate, does it? in other words, the bigger the surface area the better the accuracy of the calculus?

It shouldn't affect the calculation. The rate shouldn't affect what the empirical formula is, because you can have the same molecule undergoing the same reaction but at different rates depending on external factors like temperature and surface area which don't change it. What calculations are you using to find the empirical formula?

1. measurement of crucible weight

2. measurement of crucible weight with magnesium (3 separate crucibles: w/ Mg ribbon, w/ Mg pieces, w/ Mg powder)

3. magnesium oxidation (of the 3)

4. measurement of magnesium oxide (after letting it cool down etc.)

5. calculation of Mg moles (by subtracting the result of step 2 from that of step 1)

6. calculation of O moles (by subtracting the result of step 4 from that of step 2)

7. calculation of ratio for MgO

Mg₉O₅ (crucible w/ ribbon), Mg₁₉O₁₀ (crucible w/ pieces) and Mg₁₈O (crucible w/ powder).

reaction rate:(crucible w/ pieces)<(crucible w/ ribbon) < (crucible w/ powder).

[Emmi]

1. measurement of crucible weight

2. measurement of crucible weight with magnesium (3 separate crucibles: w/ Mg ribbon, w/ Mg pieces, w/ Mg powder)

3. magnesium oxidation (of the 3)

4. measurement of magnesium oxide (after letting it cool down etc.)

5. calculation of Mg moles (by subtracting the result of step 2 from that of step 1)

6. calculation of O moles (by subtracting the result of step 4 from that of step 2)

7. calculation of ratio for MgO

Mg₉O₅ (crucible w/ ribbon), Mg₁₉O₁₀ (crucible w/ pieces) and Mg₁₈O (crucible w/ powder).

reaction rate:(crucible w/ pieces)<(crucible w/ ribbon) < (crucible w/ powder).

Hmm...your procedure seems fine, but getting things like Mg₉O₅ and such don't really make sense. Double check your calculations, because the white powder you see after you burn magnesium in oxygen is MgO. I'm not so sure why you're getting formulas like that.

[MWL]

1. measurement of crucible weight

2. measurement of crucible weight with magnesium (3 separate crucibles: w/ Mg ribbon, w/ Mg pieces, w/ Mg powder)

3. magnesium oxidation (of the 3)

4. measurement of magnesium oxide (after letting it cool down etc.)

5. calculation of Mg moles (by subtracting the result of step 2 from that of step 1)

6. calculation of O moles (by subtracting the result of step 4 from that of step 2)

7. calculation of ratio for MgO

Mg₉O₅ (crucible w/ ribbon), Mg₁₉O₁₀ (crucible w/ pieces) and Mg₁₈O (crucible w/ powder).

reaction rate:(crucible w/ pieces)<(crucible w/ ribbon) < (crucible w/ powder).

Hmm...your procedure seems fine, but getting things like Mg₉O₅ and such don't really make sense. Double check your calculations, because the white powder you see after you burn magnesium in oxygen is MgO. I'm not so sure why you're getting formulas like that.

yeah... actually the experiment really just has a crappy accuracy...

so you said the different reaction rates don't affect the empirical formula?

and how about their different efficiencies? wouldn't the different surface areas have different levels of reaction in relation to how much of the magnesium was actually consumed?

[Emmi]

1. measurement of crucible weight

2. measurement of crucible weight with magnesium (3 separate crucibles: w/ Mg ribbon, w/ Mg pieces, w/ Mg powder)

3. magnesium oxidation (of the 3)

4. measurement of magnesium oxide (after letting it cool down etc.)

5. calculation of Mg moles (by subtracting the result of step 2 from that of step 1)

6. calculation of O moles (by subtracting the result of step 4 from that of step 2)

7. calculation of ratio for MgO

Mg₉O₅ (crucible w/ ribbon), Mg₁₉O₁₀ (crucible w/ pieces) and Mg₁₈O (crucible w/ powder).

reaction rate:(crucible w/ pieces)<(crucible w/ ribbon) < (crucible w/ powder).

Hmm...your procedure seems fine, but getting things like Mg₉O₅ and such don't really make sense. Double check your calculations, because the white powder you see after you burn magnesium in oxygen is MgO. I'm not so sure why you're getting formulas like that.

yeah... actually the experiment really just has a crappy accuracy...

so you said the different reaction rates don't affect the empirical formula?

and how about their different efficiencies? wouldn't the different surface areas have different levels of reaction in relation to how much of the magnesium was actually consumed?

Reaction rates don't affect the empirical formula. If I reacted ethene with hydrogen gas to form ethane, without a catalyst I'd still get ethane, it's just that the rate is so slow it's hard to measure. If I put in nickel as a catalyst in I'll get ethane, just at a much faster and noticeable rate.

I would presume that different surface areas have different "efficiencies," if by efficiency you mean how well it works in speeding up the reaction. Like you said earlier, in order of decreasing rate, powdered fragments < small chunks < whole chunk because of the surface areas. Just keep in mind that most reactions don't have a 100% yield to them if this is what you mean by efficiency, and this is why you have percent yields. A theoretical yield assumes that the entirety of the reactants formed into products and there were no side reactions or unreacted substance (however, these tend to occur and the actual amount you measure is your actual yield which is rarely your theoretical yield).

[MWL]

what you mean by that is that the speed of the reaction won't have an influence on how "well" it reacted because that's related to the efficiency of the theoretic yield.

on the other hand there is an influence of the theoretic yield on the experimental calculation of the formula.

therefore the speed won't have an indirect impact on the calculation fomula. right?

[Emmi]

what you mean by that is that the speed of the reaction won't have an influence on how "well" it reacted because that's related to the efficiency of the theoretic yield.

on the other hand there is an influence of the theoretic yield on the experimental calculation of the formula.

therefore the speed won't have an indirect impact on the calculation fomula. right?

I believe so, I'm but not entirely certain on that.

[MWL]

what you mean by that is that the speed of the reaction won't have an influence on how "well" it reacted because that's related to the efficiency of the theoretic yield.

on the other hand there is an influence of the theoretic yield on the experimental calculation of the formula.

therefore the speed won't have an indirect impact on the calculation fomula. right?

I believe so, I'm but not entirely certain on that.

what you mean by that is that the speed of the reaction won't have an influence on how "well" it reacted because that's related to the efficiency of the theoretic yield.

on the other hand there is an influence of the theoretic yield on the experimental calculation of the formula.

therefore the speed won't have an indirect impact on the calculation fomula. right?

I believe so, I'm but not entirely certain on that.

Thank you very much!